A force of \[50\,{\text{kgf}}\] is applied to the smaller piston of a hydraulic machine. Neglecting the friction, the force exerted on the large piston, if the diameters of the piston are \[5\,{\text{cm}}\] and \[25\,{\text{cm}}\] respectively is ‘\[X\]’ \[{\text{N}}\] . Find \[\dfrac{X}{{250}}\] .
Answer
290.4k+ views
Hint: First we will calculate the areas of the smaller and the larger pistons. Then we will calculate the pressure applied onto each of them. Eventually, we will equate the two equations and manipulate accordingly to obtain the result.
Complete step by step answer:
In the given question, we are supplied with the following data:
The diameter of the smaller piston is \[5\,{\text{cm}}\] .
The diameter of the larger piston is \[25\,{\text{cm}}\] .
The force exerted on the smaller piston is \[50\,{\text{kgf}}\] .
We are asked to find the force exerted on the larger piston.
We will apply the trick that the pressure is evenly distributed in all the pistons, as we know water is an incompressible liquid. The pressure you put on the smaller piston and the pressure on the larger piston will be equal.
We know the formula which relates pressure, force and area is given below:
\[P = \dfrac{F}{A}\] …… (1)
Where,
\[P\] indicates pressure.
\[F\] indicates force applied.
\[A\] indicates the area on which the force is applied.
Again, we will find the area of the smaller piston:
\[A = \pi {\left( {\dfrac{D}{2}} \right)^2}\] …… (2)
Where,
\[A\] indicates area.
\[D\] indicates diameter of the piston.
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
\[
A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\
A = 3.14 \times {\left( {\dfrac{5}{2}} \right)^2} \\
A = 19.6\,{\text{c}}{{\text{m}}^2} \\
\]
Therefore, the area of the smaller piston is found to be \[19.6\,{\text{c}}{{\text{m}}^2}\] .
So, the pressure can be calculated by using the equation (1), as follows:
\[P = \dfrac{F}{A}\]
\[P = \dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}}\] …… (3)
Again, we will find the area of the larger piston:
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
\[
A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\
A = 3.14 \times {\left( {\dfrac{{25}}{2}} \right)^2} \\
A = 490.6\,{\text{c}}{{\text{m}}^2} \\
\]
Therefore, the area of the smaller piston is found to be \[490.6\,{\text{c}}{{\text{m}}^2}\] .
So, the pressure can be calculated by using the equation (1), as follows:
\[P = \dfrac{F}{A}\]
\[P = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}}\] …… (4)
Now, we equate equation (3) and (4), and we get:
\[
\dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}} \\
X = \dfrac{{50\,{\text{kgf}} \times 490.6\,{\text{c}}{{\text{m}}^2}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} \\
X = 1251.51\,{\text{kgf}} \\
X \sim 1252\,{\text{kgf}} \\
\]
Hence, the force exerted on the larger piston is \[1252\,{\text{kgf}}\] .
Note:The given problem is based on the mechanical properties of fluid. While calculating the area, you can also use the S.I units, there’s no issue and the answer will come the same because the units of the area will eventually cancel each other. These types of hydraulic lifts are used to lift heavy vehicles, by applying pressure on one end.
Complete step by step answer:
In the given question, we are supplied with the following data:
The diameter of the smaller piston is \[5\,{\text{cm}}\] .
The diameter of the larger piston is \[25\,{\text{cm}}\] .
The force exerted on the smaller piston is \[50\,{\text{kgf}}\] .
We are asked to find the force exerted on the larger piston.
We will apply the trick that the pressure is evenly distributed in all the pistons, as we know water is an incompressible liquid. The pressure you put on the smaller piston and the pressure on the larger piston will be equal.
We know the formula which relates pressure, force and area is given below:
\[P = \dfrac{F}{A}\] …… (1)
Where,
\[P\] indicates pressure.
\[F\] indicates force applied.
\[A\] indicates the area on which the force is applied.
Again, we will find the area of the smaller piston:
\[A = \pi {\left( {\dfrac{D}{2}} \right)^2}\] …… (2)
Where,
\[A\] indicates area.
\[D\] indicates diameter of the piston.
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
\[
A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\
A = 3.14 \times {\left( {\dfrac{5}{2}} \right)^2} \\
A = 19.6\,{\text{c}}{{\text{m}}^2} \\
\]
Therefore, the area of the smaller piston is found to be \[19.6\,{\text{c}}{{\text{m}}^2}\] .
So, the pressure can be calculated by using the equation (1), as follows:
\[P = \dfrac{F}{A}\]
\[P = \dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}}\] …… (3)
Again, we will find the area of the larger piston:
Now, we calculate the area of the smaller piston by substituting the required values in equation (2):
\[
A = \pi {\left( {\dfrac{D}{2}} \right)^2} \\
A = 3.14 \times {\left( {\dfrac{{25}}{2}} \right)^2} \\
A = 490.6\,{\text{c}}{{\text{m}}^2} \\
\]
Therefore, the area of the smaller piston is found to be \[490.6\,{\text{c}}{{\text{m}}^2}\] .
So, the pressure can be calculated by using the equation (1), as follows:
\[P = \dfrac{F}{A}\]
\[P = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}}\] …… (4)
Now, we equate equation (3) and (4), and we get:
\[
\dfrac{{50\,{\text{kgf}}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} = \dfrac{X}{{490.6\,{\text{c}}{{\text{m}}^2}}} \\
X = \dfrac{{50\,{\text{kgf}} \times 490.6\,{\text{c}}{{\text{m}}^2}}}{{19.6\,{\text{c}}{{\text{m}}^2}}} \\
X = 1251.51\,{\text{kgf}} \\
X \sim 1252\,{\text{kgf}} \\
\]
Hence, the force exerted on the larger piston is \[1252\,{\text{kgf}}\] .
Note:The given problem is based on the mechanical properties of fluid. While calculating the area, you can also use the S.I units, there’s no issue and the answer will come the same because the units of the area will eventually cancel each other. These types of hydraulic lifts are used to lift heavy vehicles, by applying pressure on one end.
Last updated date: 04th Jun 2023
•
Total views: 290.4k
•
Views today: 5.46k
Recently Updated Pages
Which element possesses the biggest atomic radii A class 11 chemistry JEE_Main

The highly efficient method of obtaining beryllium class 11 chemistry JEE_Main

Which of the following sulphates has the highest solubility class 11 chemistry JEE_Main

Amongst the metal Be Mg Ca and Sr of group 2 of the class 11 chemistry JEE_Main

Which of the following metals is present in the greencolored class 11 chemistry JEE_Main

To prevent magnesium from oxidation in the electrolytic class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE
